Problem: $f\,^{\prime}(x)=16x^3+6x-7$ and $f(5)=2500$. $f(-1) = $
Finding $f(x)$ We have $f'(x)=16x^3+6x-7$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (16x^3+6x-7)\,dx \\\\ & = {4x^4+3x^2-7x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(5)=2500$. Here's what we get when we plug in $5$ : $\begin{aligned}f(5)&={4(5)^4+3(5)^2-7(5)} {+ C}\\\\ &={2540} {+ C} \end{aligned}$ We are given that this must equal $2500$ : $2500 = {2540} {+ C}$ Solving the equation gives us ${C=-40}$. Finding $f(-1)$ Now, we have that $f(x)={4x^4+3x^2-7x} {-40}$. Let's find $f(-1)$ by plugging in $-1$ : $\begin{aligned}f(-1)&=4(-1)^4+3(-1)^2-7(-1) - 40\\\\ &=-26 \end{aligned}$ The answer $f(-1) = -26$